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Longest Increasing Subsequence Length and Value

Below code gives the length of longest increasing subsequence. Code works fine.
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
With the help of below code I want to print which array elements are part of the subsequence i.e [2,3,7,101],
How can I do it? Or do I need to write completely different code?
Is there any generic way to do for other problems to do like longest palindromic subsequence?
public int lengthOfLIS(int[] nums) {
return helper(nums, 0, Integer.MIN_VALUE);
}
public int helper(int[] arr, int i, int prev) {
if (i == arr.length)
return 0;
int include = 0;
if (arr[i] > prev) {
include = 1 + helper(arr, i + 1, arr[i]);
}
int exclude = helper(arr, i + 1, prev);
return Math.max(include, exclude);
}

public void longestIncreasingSubsequence(int[] numbers)
{
List<Integer> result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
for (int i = 0; i < numbers.length; i++) {
temp.add(numbers[i]);
if ( i + 1 == numbers.length || numbers[i] > numbers[i + 1])
{
if(temp.size() >= result.size())
{
result.clear();
result.addAll(temp);
}
temp.clear();
}
}
System.out.println(result);
}
Try this
I got your point. Plz Check my new code
public void longestSubSequence(int[] nums)
{
int index=0;
List<Integer> result = new ArrayList<>();
while (index<nums.length)
{
int val=nums[index];
List<Integer> temp = new ArrayList<>();
temp.add(val);
for (int i = index+1; i < nums.length; i++) {
int currentVal = nums[i];
if (val < currentVal)
{
val= currentVal;
temp.add(currentVal);
}
}
if(temp.size()>= result.size())
{
result.clear();
result.addAll(temp);
temp.clear();
}
index++;
}
System.out.println(result);
}

26. Remove Duplicates from Sorted Array - Java

Question : Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
public int removeDuplicates(int[] nums) {
if (nums.length == 0) return 0;
int i = 0;
for (int j = 1; j < nums.length; j++) {
if (nums[j] != nums[i]) {
i++;
nums[i] = nums[j];
}
}
return i + 1;
}
What exactly does the return statement do here. What does return i + 1 mean here ?

The return i + 1 is returning how many unique integers are there. I believe this is a Leetcode problem, and since its in place, the int[] is passed in by reference, Leetcode wants to know how many numbers to check (you're supposed to put the unique numbers in the first i + 1 spots).
If you look at the question, it says:
Which means that you return the length of the array.
So, if you have the array [1,1,2,3,4,4], you would turn that into [1,2,3,4,...], where the ... is the rest of the array. However, you return 4 because the length of the new array should be 4.
Hope this clears things up for you!

Your question has been already answered here; in addition to that, we can also start from zero and remove the first if statement:
Test with a b.java file:
import java.util.*;
class Solution {
public static final int removeDuplicates(
final int[] nums
) {
int i = 0;
for (int num : nums)
if (i == 0 || num > nums[i - 1]) {
nums[i++] = num;
}
return i;
}
}
class b {
public static void main(String[] args) {
System.out.println(new Solution().removeDuplicates(new int[] { 1, 1, 2}));
System.out.println(new Solution().removeDuplicates(new int[] { 0, 0, 1, 1, 1, 2, 2, 3, 3, 4}));
}
}
prints
2
5

I tried in this easy way. Here Time complexity is O(n) and space
complexity: O(1).
static int removeDuplicates(int[] nums){
if(nums.length == 0) {
return 0;
}
int value = nums[0];
int lastIndex = 0;
int count = 1;
for (int i = 1; i < nums.length; i++) {
if(nums[i] > value) {
value = nums[i];
lastIndex = lastIndex+1;
nums[lastIndex] = value;
count++;
}
}
return count;
}

class Solution {
public int removeDuplicates(int[] nums) {
int n = nums.length;
if (n == 0 || n == 1)
return n;
int j = 0;
for (int i=0; i<n-1; i++)
if (nums[i]!= nums[i+1])
nums[j++] = nums[i];
nums[j++]=nums[n-1];
return j;
}
}

public class RemoveDuplicateSortedArray {
//Remove Duplicates from Sorted Array
public static void main(String[] args) {
int[] intArray = new int[]{0, 0, 1, 1, 1, 2, 2, 3, 3, 4};
int count = extracted(intArray);
for (int i = 0; i < count; i++) {
System.out.println(intArray[i]);
}
}
private static int extracted(int[] intArray) {
int size = intArray.length;
int count = 1;
if (size == 1) {
return 1;
} else if (size == 2) {
if (intArray[0] == intArray[1]) {
return 1;
} else {
return 2;
}
} else {
for (int i = 0, j = i + 1; j < size; j++) {
if (intArray[i] < intArray[j]) {
i++;
intArray[i] = intArray[j];
count++;
}
}
return count;
}
}
}

I made a MergeSort method with the firstHalf and secondHalf Method; index 0 ~ middle, middle+1 ~ end. However, this calls the StackOverFlow Error

public static void mergeSort(int[] data) {
int[] left = firstHalf(data);
int[] right = secondHalf(data);
if (data.length > 1) {
mergeSort(left);
mergeSort(right);
merge(data, left, right);
}
}
public static void merge(int[] data, int[] left, int[] right) {
int tempArraySize = data.length;
int mergedNumbers[] = new int[tempArraySize]; //Temp array to take the sorted array
int mergePos;
int leftPos;
int rightPos;
int middle = data.length / 2;
mergePos = 0;
leftPos = 0; // 0 index
rightPos = middle + 1; //j is middle index
while (leftPos <= middle && rightPos <= data.length - 1) {
if (left[leftPos] < right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
}
mergePos++;
}
// when the right half array finishes sorting
while (leftPos <= middle) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
// when the left half array finishes sorting
while (rightPos <= data.length - 1) {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
// give value to the original array
for (mergePos = 0; mergePos < tempArraySize; ++mergePos) {
data[leftPos + mergePos] = mergedNumbers[mergePos];
}
}
public static int[] firstHalf(int[] data) {
int[] tempFirst = new int[(data.length / 2) + 1];
for (int i = 0; i <= data.length / 2; i++) {
tempFirst[i] = data[i];
}
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int[] tempSecond = new int[(data.length / 2) + 1];
for (int i = (data.length / 2) + 1; i < data.length; i++) { // Middle to the end
for (int j = 0; j <= data.length / 2; j++) {
tempSecond[j] = data[i];
}
}
return tempSecond;
}
This is what I made.
My mergeSort method makes an error java.lang.StackOverflowError
What mistakes I made?
I made the firstHalf and secondHalf methods to get the index 0 ~ middle and middle+1 ~ end.
Those methods are made to get the value from the original 'data' Array.
The merge method is as same as the common MergeSort code.
Do I have to build a base case in the mergeSort method?

With this approach, it is simpler to return merged arrays. It would be faster to do a one time allocation of a temporary array, and use indexing to merge data between the two arrays rather than creating temporary arrays and copy data. Fixes noted in comments.
public static int[] mergeSort(int[] data) { // fix
int[] left = firstHalf(data);
int[] right = secondHalf(data);
if(data.length < 2) // change
return data; // fix
left = mergeSort(left); // fix
right = mergeSort(right); // fix
return merge(left, right); // fix
}
public static int[] merge(int[] left, int[] right) { // fix
int mergedNumbers [] = new int[left.length+right.length]; // fix
int mergePos = 0; // fix
int leftPos = 0; // fix
int rightPos = 0; // fix
while (leftPos < left.length && rightPos < right.length) { // fix
if (left[leftPos] < right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
}
mergePos++;
}
while (leftPos < left.length) { // fix
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
while (rightPos < right.length) { // fix
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
return mergedNumbers; // fix
}
public static int[] firstHalf(int[] data) {
int j = (data.length/2); // fix
int[] tempFirst = new int[j]; // fix
for(int i = 0; i < tempFirst.length; i++) // fix
tempFirst[i] = data[i];
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int j = (data.length/2); // fix
int[] tempSecond = new int[data.length-j]; // fix
for(int i = 0; i < tempSecond.length; i++) // fix
tempSecond[i] = data[i+j]; // fix
return tempSecond;
}
The wiki article has a somewhat optimized approach for top down merge sort:
https://en.wikipedia.org/wiki/Merge_sort#Top-down_implementation

The problems in your code come from the confusing convention to specify ranges with included boundaries. You should instead consider the upper bound to be excluded: this would avoid the numerous +1/-1 adjustments required for the included convention, some of which are inconsistent in your code:
leftHalf creates an array with length (data.length / 2) + 1, including the element at offset mid = data.length / 2. This is what the merge method expects, but is not optimal as it would make unbalanced halves for even sized arrays, and more problematically would return a 2 element slice for a 2 element array, which causes an infinite recursion and explains the Stack Overflow exception you get.
rightHalf also creates an array with length (data.length / 2) + 1, which is incorrect, the length should be array with lengthdata.length - ((data.length / 2) + 1)`, which would be an empty array for a 2 element array.
Furthermore, rightHalf uses two nested for loops to copy the values from the argument array, which is incorrect.
your index range into the right array is incorrect in the merge method.
the index into data is incorrect in data[leftPos + mergePos] = mergedNumbers[mergePos]; it should be just:
data[mergePos] = mergedNumbers[mergePos];
Here is a modified version, with a less error prone convention:
// sort elements in data in place
public static void mergeSort(int[] data) {
if (data.length > 1) {
int[] left = firstHalf(data);
int[] right = secondHalf(data);
mergeSort(left);
mergeSort(right);
merge(data, left, right);
}
}
public static void merge(int[] data, int[] left, int[] right) {
int leftLength = left.length;
int rightLength = right.length;
int length = leftLength + rightLength;
int mergedNumbers[] = new int[length]; //Temp array to received the merged array
int leftPos = 0;
int rightPos = 0;
int mergePos = 0;
while (leftPos < leftLength && rightPos < rightLength) {
if (left[leftPos] <= right[rightPos]) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
} else {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
}
// copy the remaining entries in the left half
while (leftPos < leftLength) {
mergedNumbers[mergePos] = left[leftPos];
leftPos++;
mergePos++;
}
// copy the remaining entries in the right half
while (rightPos < rightLength) {
mergedNumbers[mergePos] = right[rightPos];
rightPos++;
mergePos++;
}
// copy the values back to the original array
for (mergePos = 0; mergePos < length; mergePos++) {
data[mergePos] = mergedNumbers[mergePos];
}
}
public static int[] firstHalf(int[] data) {
int leftLength = data.length / 2;
int[] tempFirst = new int[leftLength];
for (int i = 0; i < leftLength; i++) {
tempFirst[i] = data[i];
}
return tempFirst;
}
public static int[] secondHalf(int[] data) {
int leftLength = data.length / 2;
int rightLength = data.length - leftLength;
int[] tempSecond = new int[rightLength];
for (int i = 0; i < rightLength; i++) {
tempSecond[i] = data[LeftLength + i];
}
return tempSecond;
}

sorting using heapsort in java

import java.util.Arrays;
public class Test{
//main method
public static void main(String... args){
int[] a = new int[]{16,14,10,8,7,9,3,2,4,1};
System.out.println("unsorted array " + Arrays.toString(a));
heap_sort(a);
System.out.println("Sorted array " + Arrays.toString(a));
}
//returns left child index of given index
private static int left_child(int i){
return (i * 2 + 1);
}
//returns right child index of the given index
private static int right_child(int i){
return (i * 2 + 2);
}
public static void max_heapify(int[] array , int index , int heap_size){
int largest = index;
int left = left_child(index);
int right = right_child(index);
if(left < heap_size && array[left] > array[index]){
largest = left;
}
else largest = index;
if (right < heap_size && array[right] > array[largest]){
largest = right;
}
if(largest != index){
//exchange array[index] with array[largest]
int a = array[index];
array[index] = array[largest];
array[largest] = a;
max_heapify(array,largest,heap_size);
}
}
public static void build_max_heap(int[] array){
int heap_size = array.length - 1;
for (int i = ((array.length - 1) / 2) ; i >= 0 ; i--){
max_heapify(array, i, heap_size);
}
}
//main heap sort function
public static void heap_sort(int[] array){
int heap_size = array.length - 1;
build_max_heap(array);
for(int i = (array.length - 1) ; i > 0 ; i--){
//exchange array[0] with array[i]
int a = array[0];
array[0] = array[i];
array[i] = a;
heap_size = heap_size - 1;
max_heapify(array , 1 , heap_size);
}
}
}
I have tried almost every possibility but it doesn't works fine .Could u find out where i am wrong.
Output of my code is :
unsorted array [16, 14, 10, 8, 7, 9, 3, 2, 4, 1]
Sorted array [14, 10, 8, 7, 9, 3, 2, 4, 1, 16]

There are some problems:
- Left and right child are modified because you loose the first element in the array.
- In max_heapify function, the loops are <=
- In heap_sort function you must pass 0 instead of 1 to the max_heapify function.
//returns left child index of given index
private static int left_child(int i)
{
return (i * 2);
}
//returns right child index of the given index
private static int right_child(int i)
{
return (i * 2 + 1);
}
public static void max_heapify(int[] array, int index, int heap_size)
{
int largest = index;
int left = left_child(index);
int right = right_child(index);
if (left <= heap_size && array[left] > array[index])
{
largest = left;
}
if (right <= heap_size && array[right] > array[largest])
{
largest = right;
}
if (largest != index)
{
//exchange array[index] with array[largest]
int a = array[index];
array[index] = array[largest];
array[largest] = a;
max_heapify(array, largest, heap_size);
}
}
public static void build_max_heap(int[] array)
{
int heap_size = array.length - 1;
for (int i = ((array.length - 1) / 2); i >= 0; i--)
{
max_heapify(array, i, heap_size);
}
}
//main heap sort function
public static void heap_sort(int[] array)
{
int heap_size = array.Length - 1;
build_max_heap(array);
for (int i = (array.Length - 1); i > 0; i--)
{
//exchange array[0] with array[i]
int a = array[0];
array[0] = array[i];
array[i] = a;
heap_size = heap_size - 1;
max_heapify(array, 0, heap_size);
}
}

The problem is that you loose items because the left and right child must be:
//returns left child index of given index
private static int left_child(int i){
return (i * 2);
}
//returns right child index of the given index
private static int right_child(int i){
return (i * 2 + 1);
}

How to merge two sorted arrays into a sorted array? [closed]

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Want to improve this question? Update the question so it focuses on one problem only by editing this post.
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Improve this question
This was asked of me in an interview and this is the solution I provided:
public static int[] merge(int[] a, int[] b) {
int[] answer = new int[a.length + b.length];
int i = 0, j = 0, k = 0;
while (i < a.length && j < b.length)
{
if (a[i] < b[j])
{
answer[k] = a[i];
i++;
}
else
{
answer[k] = b[j];
j++;
}
k++;
}
while (i < a.length)
{
answer[k] = a[i];
i++;
k++;
}
while (j < b.length)
{
answer[k] = b[j];
j++;
k++;
}
return answer;
}
Is there a more efficient way to do this?
Edit: Corrected length methods.

public static int[] merge(int[] a, int[] b) {
int[] answer = new int[a.length + b.length];
int i = 0, j = 0, k = 0;
while (i < a.length && j < b.length)
answer[k++] = a[i] < b[j] ? a[i++] : b[j++];
while (i < a.length)
answer[k++] = a[i++];
while (j < b.length)
answer[k++] = b[j++];
return answer;
}
Is a little bit more compact but exactly the same!

I'm surprised no one has mentioned this much more cool, efficient and compact implementation:
public static int[] merge(int[] a, int[] b) {
int[] answer = new int[a.length + b.length];
int i = a.length - 1, j = b.length - 1, k = answer.length;
while (k > 0)
answer[--k] =
(j < 0 || (i >= 0 && a[i] >= b[j])) ? a[i--] : b[j--];
return answer;
}
Points of Interests
Notice that it does same or less number of operations as any other O(n) algorithm but in literally single statement in a single while loop!
If two arrays are of approximately same size then constant for O(n) is same. However if arrays are really imbalanced then versions with System.arraycopy would win because internally it can do this with single x86 assembly instruction.
Notice a[i] >= b[j] instead of a[i] > b[j]. This guarantees "stability" that is defined as when elements of a and b are equal, we want elements from a before b.

A minor improvement, but after the main loop, you could use System.arraycopy to copy the tail of either input array when you get to the end of the other. That won't change the O(n) performance characteristics of your solution, though.

Any improvements that could be made would be micro-optimizations, the overall algorithm is correct.

This solution also very similar to other posts except that it uses System.arrayCopy to copy the remaining array elements.
private static int[] sortedArrayMerge(int a[], int b[]) {
int result[] = new int[a.length +b.length];
int i =0; int j = 0;int k = 0;
while(i<a.length && j <b.length) {
if(a[i]<b[j]) {
result[k++] = a[i];
i++;
} else {
result[k++] = b[j];
j++;
}
}
System.arraycopy(a, i, result, k, (a.length -i));
System.arraycopy(b, j, result, k, (b.length -j));
return result;
}

Here is updated function. It removes duplicates, hopefully someone will find this usable:
public static long[] merge2SortedAndRemoveDublicates(long[] a, long[] b) {
long[] answer = new long[a.length + b.length];
int i = 0, j = 0, k = 0;
long tmp;
while (i < a.length && j < b.length) {
tmp = a[i] < b[j] ? a[i++] : b[j++];
for ( ; i < a.length && a[i] == tmp; i++);
for ( ; j < b.length && b[j] == tmp; j++);
answer[k++] = tmp;
}
while (i < a.length) {
tmp = a[i++];
for ( ; i < a.length && a[i] == tmp; i++);
answer[k++] = tmp;
}
while (j < b.length) {
tmp = b[j++];
for ( ; j < b.length && b[j] == tmp; j++);
answer[k++] = tmp;
}
return Arrays.copyOf(answer, k);
}

It can be done in 4 statements as below
int a[] = {10, 20, 30};
int b[]= {9, 14, 11};
int res[]=new int[a.legth+b.length];
System.arraycopy(a,0, res, 0, a.length);
System.arraycopy(b,0,res,a.length, b.length);
Array.sort(res)

GallopSearch Merge: O(log(n)*log(i)) rather than O(n)
I went ahead and implemented greybeard suggestion in the comments. Mostly because I needed a highly efficient mission critical version of this code.
The code uses a gallopSearch which is O(log(i)) where i is the
distance from the current index the relevant index exists.
The code uses a binarySearch for after the gallop search has
identified the proper,range. Since gallop limited this to a smaller
range the resulting binarySearch is also O(log(i))
The gallop and merge are performed backwards. This doesn't seem
mission critical but it allows in place merging of arrays. If one of
your arrays has enough room to store the results values, you can
simply use it as the merging array and the results array. You must specify the valid range within the array in such a case.
It does not require memory allocation in that case (big savings in critical operations). It simply makes sure it doesn't and cannot overwrite any unprocessed values (which can only be done backwards). In fact, you use the same array for both of the inputs and the results. It will suffer no ill effects.
I consistently used Integer.compare() so this could be switched out for other purposes.
There's some chance I might have goofed a little and not utilized information I have previously proven. Such as binary searching into a range of two values, for which one value was already checked. There might also be a better way to state the main loop, the flipping c value wouldn't be needed if they were combined into two operations in sequence. Since you know you will do one then the other everytime. There's room for for some polish.
This should be the most efficient way to do this, with time complexity of O(log(n)*log(i)) rather than O(n). And worst case time complexity of O(n). If your arrays are clumpy and have long strings of values together, this will dwarf any other way to do it, otherwise it'll just be better than them.
It has two read values at the ends of the merging array and the write value within the results array. After finding out which is end value is less, it does a gallop search into that array. 1, 2, 4, 8, 16, 32, etc. When it finds the range where the the other array's read value is bigger. It binary searches into that range (cuts the range in half, search the correct half, repeat until single value). Then it array copies those values into the write position. Keeping in mind that the copy is, by necessity, moved such that it cannot overwrite the same values from the either reading array (which means the write array and read array can be the same). It then performs the same operation for the other array which is now known to be less than the new read value of the other array.
static public int gallopSearch(int current, int[] array, int v) {
int d = 1;
int seek = current - d;
int prevIteration = seek;
while (seek > 0) {
if (Integer.compare(array[seek], v) <= 0) {
break;
}
prevIteration = seek;
d <<= 1;
seek = current - d;
if (seek < 0) {
seek = 0;
}
}
if (prevIteration != seek) {
seek = binarySearch(array, seek, prevIteration, v);
seek = seek >= 0 ? seek : ~seek;
}
return seek;
}
static public int binarySearch(int[] list, int fromIndex, int toIndex, int v) {
int low = fromIndex;
int high = toIndex - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
int midVal = list[mid];
int cmp = Integer.compare(midVal, v);
if (cmp < 0) {
low = mid + 1;
} else if (cmp > 0) {
high = mid - 1;
} else {
return mid;// key found
}
}
return -(low + 1);// key not found.
}
static public int[] sortedArrayMerge(int[] a, int[] b) {
return sortedArrayMerge(null, a, a.length, b, b.length);
}
static public int[] sortedArrayMerge(int[] results, int[] a, int aRead, int b[], int bRead) {
int write = aRead + bRead, length, gallopPos;
if ((results == null) || (results.length < write)) {
results = new int[write];
}
if (aRead > 0 && bRead > 0) {
int c = Integer.compare(a[aRead - 1], b[bRead - 1]);
while (aRead > 0 && bRead > 0) {
switch (c) {
default:
gallopPos = gallopSearch(aRead, a, b[bRead-1]);
length = (aRead - gallopPos);
write -= length;
aRead = gallopPos;
System.arraycopy(a, gallopPos--, results, write, length);
c = -1;
break;
case -1:
gallopPos = gallopSearch(bRead, b, a[aRead-1]);
length = (bRead - gallopPos);
write -= length;
bRead = gallopPos;
System.arraycopy(b, gallopPos--, results, write, length);
c = 1;
break;
}
}
}
if (bRead > 0) {
if (b != results) {
System.arraycopy(b, 0, results, 0, bRead);
}
} else if (aRead > 0) {
if (a != results) {
System.arraycopy(a, 0, results, 0, aRead);
}
}
return results;
}
This should be the most efficient way to do it.
Some answers had a duplicate remove ability. That'll require an O(n) algorithm because you must actually compare each item. So here's a stand-alone for that, to be applied after the fact. You can't gallop through multiple entries all the way through if you need to look at all of them, though you could gallop through the duplicates, if you had a lot of them.
static public int removeDuplicates(int[] list, int size) {
int write = 1;
for (int read = 1; read < size; read++) {
if (list[read] == list[read - 1]) {
continue;
}
list[write++] = list[read];
}
return write;
}
Update: Previous answer, not horrible code but clearly inferior to the above.
Another needless hyper-optimization. It not only invokes arraycopy for the end bits, but also for the beginning. Processing any introductory non-overlap in O(log(n)) by a binarySearch into the data. O(log(n) + n) is O(n) and in some cases the effect will be pretty pronounced especially things like where there is no overlap between the merging arrays at all.
private static int binarySearch(int[] array, int low, int high, int v) {
high = high - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
int midVal = array[mid];
if (midVal > v)
low = mid + 1;
else if (midVal < v)
high = mid - 1;
else
return mid; // key found
}
return low;//traditionally, -(low + 1); // key not found.
}
private static int[] sortedArrayMerge(int a[], int b[]) {
int result[] = new int[a.length + b.length];
int k, i = 0, j = 0;
if (a[0] > b[0]) {
k = i = binarySearch(b, 0, b.length, a[0]);
System.arraycopy(b, 0, result, 0, i);
} else {
k = j = binarySearch(a, 0, a.length, b[0]);
System.arraycopy(a, 0, result, 0, j);
}
while (i < a.length && j < b.length) {
result[k++] = (a[i] < b[j]) ? a[i++] : b[j++];
}
if (j < b.length) {
System.arraycopy(b, j, result, k, (b.length - j));
} else {
System.arraycopy(a, i, result, k, (a.length - i));
}
return result;
}

I had to write it in javascript, here it is:
function merge(a, b) {
var result = [];
var ai = 0;
var bi = 0;
while (true) {
if ( ai < a.length && bi < b.length) {
if (a[ai] < b[bi]) {
result.push(a[ai]);
ai++;
} else if (a[ai] > b[bi]) {
result.push(b[bi]);
bi++;
} else {
result.push(a[ai]);
result.push(b[bi]);
ai++;
bi++;
}
} else if (ai < a.length) {
result.push.apply(result, a.slice(ai, a.length));
break;
} else if (bi < b.length) {
result.push.apply(result, b.slice(bi, b.length));
break;
} else {
break;
}
}
return result;
}

Apache collections supports collate method since version 4; you can do this using the collate method in:
org.apache.commons.collections4.CollectionUtils
Here quote from javadoc:
collate(Iterable<? extends O> a, Iterable<? extends O> b, Comparator<? super O> c)
Merges two sorted Collections, a and b, into a single,
sorted List such that the ordering of the elements according to
Comparator c is retained.
Do not re-invent the wheel! Document reference:
http://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/CollectionUtils.html

Here's a shortened form written in javascript:
function sort( a1, a2 ) {
var i = 0
, j = 0
, l1 = a1.length
, l2 = a2.length
, a = [];
while( i < l1 && j < l2 ) {
a1[i] < a2[j] ? (a.push(a1[i]), i++) : (a.push( a2[j]), j++);
}
i < l1 && ( a = a.concat( a1.splice(i) ));
j < l2 && ( a = a.concat( a2.splice(j) ));
return a;
}

public class Merge {
// stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
public static void merge(Comparable[] a, Comparable[] aux, int lo, int mid, int hi) {
// precondition: a[lo .. mid] and a[mid+1 .. hi] are sorted subarrays
assert isSorted(a, lo, mid);
assert isSorted(a, mid+1, hi);
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = a[k];
}
// merge back to a[]
int i = lo, j = mid+1;
for (int k = lo; k <= hi; k++) {
if (i > mid) a[k] = aux[j++];
else if (j > hi) a[k] = aux[i++];
else if (less(aux[j], aux[i])) a[k] = aux[j++];
else a[k] = aux[i++];
}
// postcondition: a[lo .. hi] is sorted
assert isSorted(a, lo, hi);
}
// mergesort a[lo..hi] using auxiliary array aux[lo..hi]
private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi) {
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, aux, lo, mid);
sort(a, aux, mid + 1, hi);
merge(a, aux, lo, mid, hi);
}
public static void sort(Comparable[] a) {
Comparable[] aux = new Comparable[a.length];
sort(a, aux, 0, a.length-1);
assert isSorted(a);
}
/***********************************************************************
* Helper sorting functions
***********************************************************************/
// is v < w ?
private static boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
// exchange a[i] and a[j]
private static void exch(Object[] a, int i, int j) {
Object swap = a[i];
a[i] = a[j];
a[j] = swap;
}
/***********************************************************************
* Check if array is sorted - useful for debugging
***********************************************************************/
private static boolean isSorted(Comparable[] a) {
return isSorted(a, 0, a.length - 1);
}
private static boolean isSorted(Comparable[] a, int lo, int hi) {
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
/***********************************************************************
* Index mergesort
***********************************************************************/
// stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
private static void merge(Comparable[] a, int[] index, int[] aux, int lo, int mid, int hi) {
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = index[k];
}
// merge back to a[]
int i = lo, j = mid+1;
for (int k = lo; k <= hi; k++) {
if (i > mid) index[k] = aux[j++];
else if (j > hi) index[k] = aux[i++];
else if (less(a[aux[j]], a[aux[i]])) index[k] = aux[j++];
else index[k] = aux[i++];
}
}
// return a permutation that gives the elements in a[] in ascending order
// do not change the original array a[]
public static int[] indexSort(Comparable[] a) {
int N = a.length;
int[] index = new int[N];
for (int i = 0; i < N; i++)
index[i] = i;
int[] aux = new int[N];
sort(a, index, aux, 0, N-1);
return index;
}
// mergesort a[lo..hi] using auxiliary array aux[lo..hi]
private static void sort(Comparable[] a, int[] index, int[] aux, int lo, int hi) {
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, index, aux, lo, mid);
sort(a, index, aux, mid + 1, hi);
merge(a, index, aux, lo, mid, hi);
}
// print array to standard output
private static void show(Comparable[] a) {
for (int i = 0; i < a.length; i++) {
StdOut.println(a[i]);
}
}
// Read strings from standard input, sort them, and print.
public static void main(String[] args) {
String[] a = StdIn.readStrings();
Merge.sort(a);
show(a);
}
}

I think introducing the skip list for the larger sorted array can reduce the number of comparisons and can speed up the process of copying into the third array. This can be good if the array is too huge.

public int[] merge(int[] a, int[] b) {
int[] result = new int[a.length + b.length];
int aIndex, bIndex = 0;
for (int i = 0; i < result.length; i++) {
if (aIndex < a.length && bIndex < b.length) {
if (a[aIndex] < b[bIndex]) {
result[i] = a[aIndex];
aIndex++;
} else {
result[i] = b[bIndex];
bIndex++;
}
} else if (aIndex < a.length) {
result[i] = a[aIndex];
aIndex++;
} else {
result[i] = b[bIndex];
bIndex++;
}
}
return result;
}

public static int[] merge(int[] a, int[] b) {
int[] mergedArray = new int[(a.length + b.length)];
int i = 0, j = 0;
int mergedArrayIndex = 0;
for (; i < a.length || j < b.length;) {
if (i < a.length && j < b.length) {
if (a[i] < b[j]) {
mergedArray[mergedArrayIndex] = a[i];
i++;
} else {
mergedArray[mergedArrayIndex] = b[j];
j++;
}
} else if (i < a.length) {
mergedArray[mergedArrayIndex] = a[i];
i++;
} else if (j < b.length) {
mergedArray[mergedArrayIndex] = b[j];
j++;
}
mergedArrayIndex++;
}
return mergedArray;
}

Algorithm could be enhanced in many ways. For instance, it is reasonable to check, if a[m-1]<b[0] or b[n-1]<a[0].
In any of those cases, there is no need to do more comparisons.
Algorithm could just copy source arrays in the resulting one in the right order.
More complicated enhancements may include searching for interleaving parts and run merge algorithm for them only.
It could save up much time, when sizes of merged arrays differ in scores of times.

This problem is related to the mergesort algorithm, in which two sorted sub-arrays are combined into a single sorted sub-array. The CLRS book gives an example of the algorithm and cleans up the need for checking if the end has been reached by adding a sentinel value (something that compares and "greater than any other value") to the end of each array.
I wrote this in Python, but it should translate nicely to Java too:
def func(a, b):
class sentinel(object):
def __lt__(*_):
return False
ax, bx, c = a[:] + [sentinel()], b[:] + [sentinel()], []
i, j = 0, 0
for k in range(len(a) + len(b)):
if ax[i] < bx[j]:
c.append(ax[i])
i += 1
else:
c.append(bx[j])
j += 1
return c

You could use 2 threads to fill the resulting array, one from front, one from back.
This can work without any synchronization in the case of numbers, e.g. if each thread inserts half of the values.

//How to merge two sorted arrays into a sorted array without duplicates?
//simple C Coding
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
main()
{
int InputArray1[] ={1,4,5,7,8,9,12,13,14,17,40};
int InputArray2[] ={4,5,11,14,15,17,18,19,112,122,122,122,122};
int n=10;
int OutputArray[30];
int i=0,j=0,k=0;
//k=OutputArray
while(i<11 && j<13)
{
if(InputArray1[i]<InputArray2[j])
{
if (k == 0 || InputArray1[i]!= OutputArray[k-1])
{
OutputArray[k++] = InputArray1[i];
}
i=i+1;
}
else if(InputArray1[i]>InputArray2[j])
{
if (k == 0 || InputArray2[j]!= OutputArray[k-1])
{
OutputArray[k++] = InputArray2[j];
}
j=j+1;
}
else
{
if (k == 0 || InputArray1[i]!= OutputArray[k-1])
{
OutputArray[k++] = InputArray1[i];
}
i=i+1;
j=j+1;
}
};
while(i<11)
{
if(InputArray1[i]!= OutputArray[k-1])
OutputArray[k++] = InputArray1[i++];
else
i++;
}
while(j<13)
{
if(InputArray2[j]!= OutputArray[k-1])
OutputArray[k++] = InputArray2[j++];
else
j++;
}
for(i=0; i<k; i++)
{
printf("sorted data:%d\n",OutputArray[i]);
};
}

var arrCombo = function(arr1, arr2){
return arr1.concat(arr2).sort(function(x, y) {
return x - y;
});
};

My favorite programming language is JavaScript
function mergeSortedArrays(a, b){
var result = [];
var sI = 0;
var lI = 0;
var smallArr;
var largeArr;
var temp;
if(typeof b[0] === 'undefined' || a[0]<b[0]){
smallArr = a;
largeArr = b;
} else{
smallArr = b;
largeArr = a;
}
while(typeof smallArr[sI] !== 'undefined'){
result.push(smallArr[sI]);
sI++;
if(smallArr[sI]>largeArr[lI] || typeof smallArr[sI] === 'undefined'){
temp = smallArr;
smallArr = largeArr;
largeArr = temp;
temp = sI;
sI = lI;
lI = temp;
}
}
return result;
}

Maybe use System.arraycopy
public static byte[] merge(byte[] first, byte[] second){
int len = first.length + second.length;
byte[] full = new byte[len];
System.arraycopy(first, 0, full, 0, first.length);
System.arraycopy(second, 0, full, first.length, second.length);
return full;
}

public static void main(String[] args) {
int[] arr1 = {2,4,6,8,10,999};
int[] arr2 = {1,3,5,9,100,1001};
int[] arr3 = new int[arr1.length + arr2.length];
int temp = 0;
for (int i = 0; i < (arr3.length); i++) {
if(temp == arr2.length){
arr3[i] = arr1[i-temp];
}
else if (((i-temp)<(arr1.length)) && (arr1[i-temp] < arr2[temp])){
arr3[i] = arr1[i-temp];
}
else{
arr3[i] = arr2[temp];
temp++;
}
}
for (int i : arr3) {
System.out.print(i + ", ");
}
}
Output is :
1, 2, 3, 4, 5, 6, 8, 9, 10, 100, 999, 1001,

You can use ternary operators for making the code a bit more compact
public static int[] mergeArrays(int[] a1, int[] a2) {
int[] res = new int[a1.length + a2.length];
int i = 0, j = 0;
while (i < a1.length && j < a2.length) {
res[i + j] = a1[i] < a2[j] ? a1[i++] : a2[j++];
}
while (i < a1.length) {
res[i + j] = a1[i++];
}
while (j < a2.length) {
res[i + j] = a2[j++];
}
return res;
}

public static int[] mergeSorted(int[] left, int[] right) {
System.out.println("merging " + Arrays.toString(left) + " and " + Arrays.toString(right));
int[] merged = new int[left.length + right.length];
int nextIndexLeft = 0;
int nextIndexRight = 0;
for (int i = 0; i < merged.length; i++) {
if (nextIndexLeft >= left.length) {
System.arraycopy(right, nextIndexRight, merged, i, right.length - nextIndexRight);
break;
}
if (nextIndexRight >= right.length) {
System.arraycopy(left, nextIndexLeft, merged, i, left.length - nextIndexLeft);
break;
}
if (left[nextIndexLeft] <= right[nextIndexRight]) {
merged[i] = left[nextIndexLeft];
nextIndexLeft++;
continue;
}
if (left[nextIndexLeft] > right[nextIndexRight]) {
merged[i] = right[nextIndexRight];
nextIndexRight++;
continue;
}
}
System.out.println("merged : " + Arrays.toString(merged));
return merged;
}
Just a small different from the original solution

To marge two sorted array in O(m+n) time complexity use below approach with one loop only.
m and n is length of first array and second array.
public class MargeSortedArray {
public static void main(String[] args) {
int[] array = new int[]{1,3,4,7};
int[] array2 = new int[]{2,5,6,8,12,45};
int[] newarry = margeToSortedArray(array, array2);
//newarray is marged array
}
// marge two sorted array with o(a+n) time complexity
public static int[] margeToSortedArray(int[] array, int[] array2) {
int newarrlen = array.length+array2.length;
int[] newarr = new int[newarrlen];
int pos1=0,pos2=0;
int len1=array.length, len2=array2.length;
for(int i =0;i<newarrlen;i++) {
if(pos1>=len1) {
newarr[i]=array2[pos2];
pos2++;
continue;
}
if(pos2>=len2) {
newarr[i]=array[pos1];
pos1++;
continue;
}
if(array[pos1]>array2[pos2]) {
newarr[i]=array2[pos2];
pos2++;
} else {
newarr[i]=array[pos1];
pos1++;
}
}
return newarr;
}
}

var arr1 = [2,10,20,30,100];
var arr2 = [2,4,5,6,7,8,9];
var j = 0;
var i =0;
var newArray = [];
for(var x=0;x< (arr1.length + arr2.length);x++){
if(arr1[i] >= arr2[j]){ //check if element arr2 is equal and less than arr1 element
newArray.push(arr2[j]);
j++;
}else if(arr1[i] < arr2[j]){ //check if element arr1 index value is less than arr2 element
newArray.push(arr1[i]);
i++;
}
else if(i == arr1.length || j < arr2.length){ // add remaining arr2 element
newArray.push(arr2[j]);
j++
}else{ // add remaining arr1 element
newArray.push(arr1[i]);
i++
}
}
console.log(newArray);

Since the question doesn't assume any specific language. Here is the solution in Python.
Assuming the arrays are already sorted.
Approach 1 - using numpy arrays:
import numpy
arr1 = numpy.asarray([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 15, 55])
arr2 = numpy.asarray([11, 32, 43, 45, 66, 76, 88])
array = numpy.concatenate((arr1,arr2), axis=0)
array.sort()
Approach 2 - Using list, assuming lists are sorted.
list_new = list1.extend(list2)
list_new.sort()

Here is my java implementation that remove duplicate.
public static int[] mergesort(int[] a, int[] b) {
int[] c = new int[a.length + b.length];
int i = 0, j = 0, k = 0, duplicateCount = 0;
while (i < a.length || j < b.length) {
if (i < a.length && j < b.length) {
if (a[i] == b[j]) {
c[k] = a[i];
i++;j++;duplicateCount++;
} else {
c[k] = a[i] < b[j] ? a[i++] : b[j++];
}
} else if (i < a.length) {
c[k] = a[i++];
} else if (j < a.length) {
c[k] = b[j++];
}
k++;
}
return Arrays.copyOf(c, c.length - duplicateCount);
}

import java.util.Arrays;
public class MergeTwoArrays {
static int[] arr1=new int[]{1,3,4,5,7,7,9,11,13,15,17,19};
static int[] arr2=new int[]{2,4,6,8,10,12,14,14,16,18,20,22};
public static void main(String[] args){
int FirstArrayLocation =0 ;
int SecondArrayLocation=0;
int[] mergeArr=new int[arr1.length + arr2.length];
for ( int i=0; i<= arr1.length + arr2.length; i++){
if (( FirstArrayLocation < arr1.length ) && (SecondArrayLocation < arr2.length)){
if ( arr1[FirstArrayLocation] <= arr2[SecondArrayLocation]){
mergeArr[i]=arr1[FirstArrayLocation];
FirstArrayLocation++;
}else{
mergeArr[i]=arr2[SecondArrayLocation];
SecondArrayLocation++;
}
}
else if(SecondArrayLocation < arr2.length){
mergeArr[i]=arr2[SecondArrayLocation];
SecondArrayLocation++;
}else if ( FirstArrayLocation < arr1.length ){
mergeArr[i]=arr1[FirstArrayLocation];
FirstArrayLocation++;
}
}
}
}